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25 September, 01:37

Matt and Anna Killian are frequent fliers on Fast-n-Go Airlines. They often fly between two cities that are a distance of 1980 miles apart. On one particular trip, they flew into the wind and the trip took 5.5 hours. The return trip with the wind behind them, only took about 4.5 hours. Find the speed of the wind and the speed of the plane in still air.

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  1. 25 September, 04:04
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    y (plane speed) 400 miles/hr

    x (wind speed) 40 miles/hr

    Step-by-step explanation:

    going into the wind make the plane flew 5,5 hours to get 1980 miles, implies the real travel speed was

    1980 / 5,5 = 360 miles/hours, and at the same time is the result of:

    speed plane (y) in direction A to B plus wind speed (x) in direction B to A. Notice opposites direction speed, so we have:

    y + (-x) = 360 (1)

    The return fly took 4,5 hours so 1980/4,5 = 440 miles / hours as the result

    Y + x = 440 (2)

    We have a two equation with two variables system, It could be solved for any of the procedures. We will use the substitution method.

    From equation (1) y + (-x) = 360 y - x = 360 y = 360 + x (1)

    In the second equation

    y + x = 440 then we replace y for its value as function of x

    (360 + x) + x = 440

    Solving for x 360 + 2x = 440 or 2x = 440 - 360

    x (440 - 360) / 2

    x = 40 miles/hr

    X (wind speed) = 40 miles/hr

    And replacing the value of x in eq. (1) y = 360 + 40 = 400

    Y (plane speed) = 400 miles / hr
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